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Question

Let f(x)=x2+6x+c, cR. If f(f(x))=0 has exactly three distinct real roots, then the value of c can be

A
9
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B
3
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C
11132
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D
11+132
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Solution

The correct option is C 11132
f(x)=x2+6x+c, cR
Since, f(x) is a polynomial of degree 2,
hence, f(f(x)) will be a polynomial of degree 4.

We know that, complex roots always occur in conjugate pair when co-efficients are real.
Since, f(f(x))=0 has three distinct real roots, so one root must be repeated.

Let roots of f(x)=0 be x=α,β
So, f(f(x))=0 has exactly three distinct real roots when α=D4a and β>D4a
α=(364c)4=c9

Now, f(D4a)=0
f(c9)=0
(c9)2+6(c9)+c=0
c211c+27=0
c=11±132

For f(x)=x2+6x+c=0, we have
α+β=6
β=3c (α=c9)

But, β>D4a
3c>c9
2c<12
c<6
c=11132

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